Consider $$h(z)=\sum_{n=1}^{\infty}\frac{(z-2)^n}{n}.$$
I wish to find an expression for $h$ as an elementary function.

This question has me stumped. I considered another function, $$f(z)=\sum_{n=1}^{\infty} n(z-2)^n.$$ This is much easier to express as an elementary function, as
$$f(z)=\sum_{n=1}^{\infty} n(z-2)^n=(z-2)\frac{d}{dz}\sum_{n=1}^{\infty} (z-2)^n=\frac{z+3}{(z+2)^2}.$$ But for the function $h$, I cannot see a similar technique or a manipulation to yield such a function.

3 Answers
3

Hint: formally $h'(z)=\sum (z-2)^{n-1}$. Calculate this sum and integrate. The answer is $Log (3-z)$ for $|z-2| <1$ where $Log$ is the principle branch of logarithm. You have to have some knowledge of logarithms in the complex plane to answer this question.